You have found the following ages (in years) of all 6 porcupines at your local zoo: $ 11,\enspace 13,\enspace 6,\enspace 1,\enspace 3,\enspace 8$ What is the average age of the porcupines at your zoo? What is the variance? You may round your answers to the nearest tenth.
Solution: Because we have data for all 6 porcupines at the zoo, we are able to calculate the population mean $({\mu})$ and population variance $({\sigma^2})$ To find the population mean , add up the values of all $6$ ages and divide by $6$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6}} $ $ {\mu} = \dfrac{11 + 13 + 6 + 1 + 3 + 8}{{6}} = {7\text{ years old}} $ Find the squared deviations from the mean for each porcupine. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $11$ years $4$ years $16$ years $^2$ $13$ years $6$ years $36$ years $^2$ $6$ years $-1$ years $1$ year $^2$ $1$ year $-6$ years $36$ years $^2$ $3$ years $-4$ years $16$ years $^2$ $8$ years $1$ year $1$ year $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{16} + {36} + {1} + {36} + {16} + {1}} {{6}} $ $ {\sigma^2} = \dfrac{{106}}{{6}} = {17.67\text{ years}^2} $ The average porcupine at the zoo is 7 years old. The population variance is 17.67 years $^2$.